Integrand size = 113, antiderivative size = 25 \[ \int \frac {e^{20 x+x \log \left (\frac {1-x \log \left (\frac {e^{-x} \log (2)}{x}\right )}{x}\right )} \left (-19-x-x^2+20 x \log \left (\frac {e^{-x} \log (2)}{x}\right )+\left (-1+x \log \left (\frac {e^{-x} \log (2)}{x}\right )\right ) \log \left (\frac {1-x \log \left (\frac {e^{-x} \log (2)}{x}\right )}{x}\right )\right )}{-1+x \log \left (\frac {e^{-x} \log (2)}{x}\right )} \, dx=e^{x \left (20+\log \left (\frac {1}{x}-\log \left (\frac {e^{-x} \log (2)}{x}\right )\right )\right )} \]
Time = 0.10 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.04 \[ \int \frac {e^{20 x+x \log \left (\frac {1-x \log \left (\frac {e^{-x} \log (2)}{x}\right )}{x}\right )} \left (-19-x-x^2+20 x \log \left (\frac {e^{-x} \log (2)}{x}\right )+\left (-1+x \log \left (\frac {e^{-x} \log (2)}{x}\right )\right ) \log \left (\frac {1-x \log \left (\frac {e^{-x} \log (2)}{x}\right )}{x}\right )\right )}{-1+x \log \left (\frac {e^{-x} \log (2)}{x}\right )} \, dx=e^{20 x} \left (\frac {1}{x}-\log \left (\frac {e^{-x} \log (2)}{x}\right )\right )^x \]
Integrate[(E^(20*x + x*Log[(1 - x*Log[Log[2]/(E^x*x)])/x])*(-19 - x - x^2 + 20*x*Log[Log[2]/(E^x*x)] + (-1 + x*Log[Log[2]/(E^x*x)])*Log[(1 - x*Log[L og[2]/(E^x*x)])/x]))/(-1 + x*Log[Log[2]/(E^x*x)]),x]
Time = 1.29 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.16, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.009, Rules used = {7257}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {e^{20 x+x \log \left (\frac {1-x \log \left (\frac {e^{-x} \log (2)}{x}\right )}{x}\right )} \left (-x^2-x+20 x \log \left (\frac {e^{-x} \log (2)}{x}\right )+\left (x \log \left (\frac {e^{-x} \log (2)}{x}\right )-1\right ) \log \left (\frac {1-x \log \left (\frac {e^{-x} \log (2)}{x}\right )}{x}\right )-19\right )}{x \log \left (\frac {e^{-x} \log (2)}{x}\right )-1} \, dx\) |
\(\Big \downarrow \) 7257 |
\(\displaystyle e^{20 x} \left (\frac {1-x \log \left (\frac {e^{-x} \log (2)}{x}\right )}{x}\right )^x\) |
Int[(E^(20*x + x*Log[(1 - x*Log[Log[2]/(E^x*x)])/x])*(-19 - x - x^2 + 20*x *Log[Log[2]/(E^x*x)] + (-1 + x*Log[Log[2]/(E^x*x)])*Log[(1 - x*Log[Log[2]/ (E^x*x)])/x]))/(-1 + x*Log[Log[2]/(E^x*x)]),x]
3.26.100.3.1 Defintions of rubi rules used
Int[(F_)^(v_)*(u_), x_Symbol] :> With[{q = DerivativeDivides[v, u, x]}, Sim p[q*(F^v/Log[F]), x] /; !FalseQ[q]] /; FreeQ[F, x]
Time = 2.71 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.08
method | result | size |
parallelrisch | \({\mathrm e}^{x \left (\ln \left (-\frac {x \ln \left (\frac {\ln \left (2\right ) {\mathrm e}^{-x}}{x}\right )-1}{x}\right )+20\right )}\) | \(27\) |
int(((x*ln(ln(2)/exp(x)/x)-1)*ln((-x*ln(ln(2)/exp(x)/x)+1)/x)+20*x*ln(ln(2 )/exp(x)/x)-x^2-x-19)*exp(x*ln((-x*ln(ln(2)/exp(x)/x)+1)/x)+20*x)/(x*ln(ln (2)/exp(x)/x)-1),x,method=_RETURNVERBOSE)
Time = 0.25 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.12 \[ \int \frac {e^{20 x+x \log \left (\frac {1-x \log \left (\frac {e^{-x} \log (2)}{x}\right )}{x}\right )} \left (-19-x-x^2+20 x \log \left (\frac {e^{-x} \log (2)}{x}\right )+\left (-1+x \log \left (\frac {e^{-x} \log (2)}{x}\right )\right ) \log \left (\frac {1-x \log \left (\frac {e^{-x} \log (2)}{x}\right )}{x}\right )\right )}{-1+x \log \left (\frac {e^{-x} \log (2)}{x}\right )} \, dx=e^{\left (x \log \left (-\frac {x \log \left (\frac {e^{\left (-x\right )} \log \left (2\right )}{x}\right ) - 1}{x}\right ) + 20 \, x\right )} \]
integrate(((x*log(log(2)/exp(x)/x)-1)*log((-x*log(log(2)/exp(x)/x)+1)/x)+2 0*x*log(log(2)/exp(x)/x)-x^2-x-19)*exp(x*log((-x*log(log(2)/exp(x)/x)+1)/x )+20*x)/(x*log(log(2)/exp(x)/x)-1),x, algorithm=\
Timed out. \[ \int \frac {e^{20 x+x \log \left (\frac {1-x \log \left (\frac {e^{-x} \log (2)}{x}\right )}{x}\right )} \left (-19-x-x^2+20 x \log \left (\frac {e^{-x} \log (2)}{x}\right )+\left (-1+x \log \left (\frac {e^{-x} \log (2)}{x}\right )\right ) \log \left (\frac {1-x \log \left (\frac {e^{-x} \log (2)}{x}\right )}{x}\right )\right )}{-1+x \log \left (\frac {e^{-x} \log (2)}{x}\right )} \, dx=\text {Timed out} \]
integrate(((x*ln(ln(2)/exp(x)/x)-1)*ln((-x*ln(ln(2)/exp(x)/x)+1)/x)+20*x*l n(ln(2)/exp(x)/x)-x**2-x-19)*exp(x*ln((-x*ln(ln(2)/exp(x)/x)+1)/x)+20*x)/( x*ln(ln(2)/exp(x)/x)-1),x)
Time = 0.35 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.12 \[ \int \frac {e^{20 x+x \log \left (\frac {1-x \log \left (\frac {e^{-x} \log (2)}{x}\right )}{x}\right )} \left (-19-x-x^2+20 x \log \left (\frac {e^{-x} \log (2)}{x}\right )+\left (-1+x \log \left (\frac {e^{-x} \log (2)}{x}\right )\right ) \log \left (\frac {1-x \log \left (\frac {e^{-x} \log (2)}{x}\right )}{x}\right )\right )}{-1+x \log \left (\frac {e^{-x} \log (2)}{x}\right )} \, dx=e^{\left (x \log \left (x^{2} + x \log \left (x\right ) - x \log \left (\log \left (2\right )\right ) + 1\right ) - x \log \left (x\right ) + 20 \, x\right )} \]
integrate(((x*log(log(2)/exp(x)/x)-1)*log((-x*log(log(2)/exp(x)/x)+1)/x)+2 0*x*log(log(2)/exp(x)/x)-x^2-x-19)*exp(x*log((-x*log(log(2)/exp(x)/x)+1)/x )+20*x)/(x*log(log(2)/exp(x)/x)-1),x, algorithm=\
\[ \int \frac {e^{20 x+x \log \left (\frac {1-x \log \left (\frac {e^{-x} \log (2)}{x}\right )}{x}\right )} \left (-19-x-x^2+20 x \log \left (\frac {e^{-x} \log (2)}{x}\right )+\left (-1+x \log \left (\frac {e^{-x} \log (2)}{x}\right )\right ) \log \left (\frac {1-x \log \left (\frac {e^{-x} \log (2)}{x}\right )}{x}\right )\right )}{-1+x \log \left (\frac {e^{-x} \log (2)}{x}\right )} \, dx=\int { -\frac {{\left (x^{2} - 20 \, x \log \left (\frac {e^{\left (-x\right )} \log \left (2\right )}{x}\right ) - {\left (x \log \left (\frac {e^{\left (-x\right )} \log \left (2\right )}{x}\right ) - 1\right )} \log \left (-\frac {x \log \left (\frac {e^{\left (-x\right )} \log \left (2\right )}{x}\right ) - 1}{x}\right ) + x + 19\right )} e^{\left (x \log \left (-\frac {x \log \left (\frac {e^{\left (-x\right )} \log \left (2\right )}{x}\right ) - 1}{x}\right ) + 20 \, x\right )}}{x \log \left (\frac {e^{\left (-x\right )} \log \left (2\right )}{x}\right ) - 1} \,d x } \]
integrate(((x*log(log(2)/exp(x)/x)-1)*log((-x*log(log(2)/exp(x)/x)+1)/x)+2 0*x*log(log(2)/exp(x)/x)-x^2-x-19)*exp(x*log((-x*log(log(2)/exp(x)/x)+1)/x )+20*x)/(x*log(log(2)/exp(x)/x)-1),x, algorithm=\
Time = 11.34 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.84 \[ \int \frac {e^{20 x+x \log \left (\frac {1-x \log \left (\frac {e^{-x} \log (2)}{x}\right )}{x}\right )} \left (-19-x-x^2+20 x \log \left (\frac {e^{-x} \log (2)}{x}\right )+\left (-1+x \log \left (\frac {e^{-x} \log (2)}{x}\right )\right ) \log \left (\frac {1-x \log \left (\frac {e^{-x} \log (2)}{x}\right )}{x}\right )\right )}{-1+x \log \left (\frac {e^{-x} \log (2)}{x}\right )} \, dx={\mathrm {e}}^{20\,x}\,{\left (x-\ln \left (\frac {\ln \left (2\right )}{x}\right )+\frac {1}{x}\right )}^x \]